If at any point fn(b)=fn−1(x), then we’re done. So assume that we get a strict increase each time up to n=|P|. Since there are only |P| elements in the entire poset, and f is monotone, fn+1(x) has to equal fn(x).
Ex 7:
For a limit ordinal α, define fα(x) as the least upper bound of fn(x) for all n<α. If α>|L|, then the set fn(x) for n<α is a set of size α that maps into a set of size L by taking the value of the element. Since there are no injections between these sets, there must be two ordinals n<m such thatfn(x)=fm(x). Since f is monotone, that implies that for every ordinal l>n, fl(x)=fn(x)and thus is a fixed point. Since n<α this proves the exercise.
Ex 8:
Starting from x, we can create a fixed point via iteration by taking α>|L|, and iterating α times as demonstrated in Ex 7. Call this fixed point fx. Suppose there was a fixed point k such that x≤k and k≤fx. Then at some point fn(x)≤fn(k)=k, but fn+1(x)≥fn+1(k)=k, which breaks the monotonicity of f unless k=fx. So fx generated this way is always the smallest fixed point greater than x.
Say we have fixed points xi. Then let x be the least upper bound of xi, and generate a fixed point from fx. So fx will be greater than each element of xi since f is monotone, and is the smallest such fixed point as shown in the above paragraph. So the poset of fixed points is semi-complete with upper bounds.
Now take our fixed points xi again. Now let x be the greatest lower bound of xi, and generate a fixed point fx. Since x≤xi and f is monotonic, fα(x)≤fα(xi)=xi, and so fx is a lower bound of xi. It has to be the greatest such bound because x itself is already the greatest such bound in our poset, and f is monotonic.
Thus the lattice of fixed points has all least upper bounds and all greatest lower bounds, and is thus complete!
Ex 6:
If at any point fn(b)=fn−1(x), then we’re done. So assume that we get a strict increase each time up to n=|P|. Since there are only |P| elements in the entire poset, and f is monotone, fn+1(x) has to equal fn(x).
Ex 7:
For a limit ordinal α, define fα(x) as the least upper bound of fn(x) for all n<α. If α>|L|, then the set fn(x) for n<α is a set of size α that maps into a set of size L by taking the value of the element. Since there are no injections between these sets, there must be two ordinals n<m such thatfn(x)=fm(x). Since f is monotone, that implies that for every ordinal l>n, fl(x)=fn(x)and thus is a fixed point. Since n<α this proves the exercise.
Ex 8:
Starting from x, we can create a fixed point via iteration by taking α>|L|, and iterating α times as demonstrated in Ex 7. Call this fixed point fx. Suppose there was a fixed point k such that x≤k and k≤fx. Then at some point fn(x)≤fn(k)=k, but fn+1(x)≥fn+1(k)=k, which breaks the monotonicity of f unless k=fx. So fx generated this way is always the smallest fixed point greater than x.
Say we have fixed points xi. Then let x be the least upper bound of xi, and generate a fixed point from fx. So fx will be greater than each element of xi since f is monotone, and is the smallest such fixed point as shown in the above paragraph. So the poset of fixed points is semi-complete with upper bounds.
Now take our fixed points xi again. Now let x be the greatest lower bound of xi, and generate a fixed point fx. Since x≤xi and f is monotonic, fα(x)≤fα(xi)=xi, and so fx is a lower bound of xi. It has to be the greatest such bound because x itself is already the greatest such bound in our poset, and f is monotonic.
Thus the lattice of fixed points has all least upper bounds and all greatest lower bounds, and is thus complete!